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In Bourbaki's General Topology, section on Filters, there is a part i find troublesome. Here is a segment from the book:

6. Filters

1. Definition of a Filter

Definition 1. A filter on a set $ X $ is a set $ \mathcal{F} $ of subsets of $ X $ which has the following properties:

  1. $ (\mathbf{F}_I) $ Every subset of $ X $ which contains a set of $ \mathcal{F} $ belongs to $ \mathcal{F} $.
  2. $ (\mathbf{F}_{II}) $ Every finite intersection of sets of $ \mathcal{F} $ belongs to $ \mathcal{F} $.
  3. $ (\mathbf{F}_{III}) $ The empty set is not in $ \mathcal{F} $.

It follows from $ (\mathbf{F}_{II}) $ and $ (\mathbf{F}_{III}) $ that every finite intersection of sets of $ \mathcal{F} $ is non-empty.

A filter $ \mathcal{F} $ on $ X $ defines a structure on $ X $, the axioms of which are $ (\mathbf{F}_I) $, $ (\mathbf{F}_{II}) $, and $ (\mathbf{F}_{III}) $; this structure is called a structure of a filtered set, and the set $ X $ endowed with this structure is called a set filtered by $ \mathcal{F} $.

Axiom $ (\mathbf{F}_{II}) $ is equivalent to the conjunction of the following two axioms:

  • $ (\mathbf{F}_{IIa}) $ The intersection of two sets of $ \mathcal{F} $ belongs to $ \mathcal{F} $.
  • $ (\mathbf{F}_{IIb}) $ $ X $ belongs to $ \mathcal{F} $.

Axioms $ (\mathbf{F}_{IIb}) $ and $ (\mathbf{F}_{III}) $ show that there is no filter on the empty set.

In order for a set of subsets which satisfies $ (\mathbf{F}_I) $ also to satisfy $ (\mathbf{F}_{IIb}) $, it is necessary and sufficient that it is not empty. A set of subsets which satisfies $ (\mathbf{F}_I) $ also satisfies $ (\mathbf{F}_{III}) $ if and only if it is different from $ \mathcal{P}(X) $.

I find it hard to see why $(\mathbf{F}_{II})$ is equivalent to $(\mathbf{F}_{IIa})$ and $(\mathbf{F}_{IIb})$. Isn't it it enough that the intersection of two sets , that belong to the filter, belongs to the filter. It can be proved with induction, without using the fact that $X$ belongs to $\mathcal{F}$, that $(\mathbf{F}_{IIa}) \implies (\mathbf{F}_{II}) $

Base Case

$n = 2$:

$$ A, B \in \mathcal{F} $$ $$(\mathbf{F}_{IIa}) \implies A \cap B \in \mathcal{F}$$

Inductive Step

$n \rightarrow n+1$:

Consider $A_1, A_2, \dots, A_n, A_{n+1} \in \mathcal{F}$. By the inductive hypothesis, we know:

$$ A_1 \cap A_2 \cap \dots \cap A_n \in \mathcal{F} $$

Now, using $(\mathbf{F}_{IIa})$ on the sets $A_1 \cap A_2 \cap \dots \cap A_n$ and $A_{n+1}$, we have:

$$ (A_1 \cap A_2 \cap \dots \cap A_n) \cap A_{n+1} \in \mathcal{F} $$

Simplifying, this gives:

$$ A_1 \cap A_2 \cap \dots \cap A_k \cap A_{k+1} \in \mathcal{F} $$

$(F_{II} \implies F_{IIa})$ is trivial, right?

So, where should the fact that $X$ belongs to $\mathcal{F}$ be used?

Milan
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  • "All finite intersections" includes the empty intersection, which is $X$. I voted to close as a duplicate - the linked question is not about filters, but it's another question about something Bourbaki writes that comes down to interpreting empty intersections properly. – Alex Kruckman Nov 05 '24 at 12:13

1 Answers1

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This is an issue with initialization: you should start at $n=0$, not $n=2$. The "empty intersection" is the whole set $X$, so $X$ is a finite intersection.

Captain Lama
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